Some tinkering around on WolframAlpha revealed that \(\displaystyle \sum_{n=0}^\infty \binom{3n}{n}x^n = \frac{2\cos(\frac{1}{3}\sin^{-1}(\frac{3\sqrt{3}}{2}\sqrt{x}))}{\sqrt{4-27x}}\). By a rather natural substitution \(x = \frac{4}{27} \sin^2 \theta\), we get